Kendall- Pelz Rule

There has been some discussion lately about alternative developers. The following rule first formulated by Kendall and later extended by Pelz enables the determination as to whether a particular organic chemical is a developer or not. In order for a chemical to be a developer it must fit the rule

A - (C=Z)n - A'

A and A' must be a chemical group - OH, -NH2, - NHR, -NRR', where R and R' are various substitution groups

C is carbon
Z is either carbon C or nitrogen N
n is 0 or any integer and represents the number of carbon-carbon or carbon-nitrogen double bonds between the groups A and A'

The rule is not perfect and there are some exceptions but works remarkably well.

Examples are,

group (a) HO - (C=C)n - OH

when n is 1 you have catechol.
when n = 2 you have hydroquinone.

group (b) HO - (C=C)n - NH2

when n = 0 you have hydrazine
when n = 2 you have paraminophenol
when n = 2 and A' is the substituted amino group - NHCH3 you have Metol.

group (c) H2N - (C=C)n - NH2 includes the phenylenediamines
group (d) HO - (C=N)n - NH2 includes the phenidones

The usual keto structural fragment of the heterocyclic ring O=C - N H must be rewritten as the enolic form HO - C=N - for the phenidones. This form shows the - OH group and the C=N bond necessary to fit the rule.

group (e) H2N - (C=N)n - NH2

Superadditivity

The combined activity of two developing agents is greater than the sum of each of them used individually.

While not part of the KP rule, for any two developing agents to exhibit superadditivity one must be of group (a) and the other from group (b) or group (d). Examples of superadditive mixtures are hydroquinone and Metol and hydroquinone and phenidone.

Taken from Mason "Photographic Processing Chemistry."

Gerald C Koch said:

group (b) HO - (C=C)n - NH2

when n = 0 you have hydrazine
when n = 2 you have paraminophenol
when n = 2 and A' is the substituted amino group - NHCH3 you have Metol.

Click to expand...

On revisiting my original post I discovered a small error in the text. The example above for n = 0 should obviously be hydroxylamine HO-NH2 and not hydrazine.

Hydrazine H2N - NH2 belongs in group (c) for n = 0.

Sorry for any confusion this may have caused.

IMO the rule is more clearly explained in Theory of the Photographic Process 3rd ed p283 (Mees & James).
I cannot draw it here but the group (C=C) should have a line sticking out the top of each C to indicate bonds making it part of a larger molecule (Often a 6 carbon ring or multiple thereof).

When drawing structural formulas in organic chemistry it is often the backbone of the molecule that is important and not the hydrogens. Such is the case for the KP rule. For example ring structures such as benzene are usually draw without the hydrogen indicated. What is important are the positions of the double bonds.

Sorry for my silly question: To my best knowledge, C=C means a double bond between C and C, like in butene and the likes. Vaguely remembering high school chemistry, I think benzene does not contain such a double bond, instead its six excess electrons for some funny orbital, hence neither hydroquinone, nor catechol have any real C=C in their structure. If that is correct, how does this Kendall-Pelz rule really work? What did I miss?

There really are no silly questions. There are two ways of thinking about the "extra"electrons in the banzene ring. Quantum mechanics considers them to be smeared about the entire ring. Unfortunately such a concept has little in the way of predictive value. An emperical model envolves a concept called resonance. Using resonance you can think of benzene as having 3 double bonds between carbons 1&2, 3&4, and 5&6. A form having bonds between carbons 2&3, 4&5, and 6&1 is also perfectly valid. The molecule is thought of as a superposition of all the possible forms, ie the molecule constantly switching between the two foms. When you add two hydroxyl groups to the ring the electron density in the ring is distorted. In the 1,2 isomer and the 1,4 isomer the OH groups are more active than when they are in the 2,3 isomer. This can easily be shown using resonance. The KP rule is based on this observation.

Resonce can be used to predict other things such as the magnitude and direction of the dipole moment of azulene an isomer of napthalene.

I hope this helps as I am condensing a semester of graduate organic chemistry into a single paragraph.

Yes, C=C symbolizes a double bond between two carbon atoms. Benzene, which has 6 carbons in a connected in a ring, is often drawn with alternating single and double bonds between all the carbons in the ring, such as this, but with the ends connected from the first carbon to the last, -C=C-C=C-C=C-

This is called "conjugation". It represents more than just a simple double bond between adjacent atoms - it denotes that electrons are being shared among the atoms, and the electrons are then free to "run around the ring", so to speak, and not be trapped inside a bond between specific adjacent atoms. Another way this is sometimes drawn is to draw all the carbons in the ring as having single bonds between them, and then on the inside of the ring, a dashed line is drawn all the way around. The dashed-line represents the conjugation - the sharing of the electrons from the double bonds with the single bonds. Some people like to think of these bonds as having 1.5 electrons in them.

Most certainly benzene, hydroquinone, and catechol have carbon-carbon double bonds (C=C), not only that, they are conjugated double bonds. They are by definition aromatic compounds, all sharing this property of having alternating carbon-carbon double and single bonds in a ring stucture.

Keep in mind this conjugation can extend into multiple rings, or even conjugated carbons connected to a ring, or multiple rings, and then more conjugated carbons. Or the reverse where there is an aromatic group connected with a string of conjugated carbons and then another aromatic group is on the opposite end. This structure is often used in photographic sensitizing dyes. Check out the structure of some organic dyes with these propterties. And keep in ming that nitrogens and oxygens can be incorporated into the conjugation.

Here's a link to cyanine dyes as used in photography:
http://en.wikipedia.org/wiki/Cyanine

And it's the conjugation in these developer compounds that allow them to spread the charge of the electon about the surface of the developer molecule that is at the heart of the subject here with the Kendal-Pelz rule.


See these pages for more info, if you like:
http://en.wikipedia.org/wiki/Conjugated_system
http://en.wikipedia.org/wiki/Organic_chemistry

Thanks a lot for the insights! From the little I remember, a classic double bond reacts quite differently from a benzene ring. It appears that the A-(C=C)n-A' notation is a strongly simplified version of Kendall and Pelzs original research work, and just looking at A-(C=C)n-A' does not tell me imbecile, why hydroquinone and catchechol are devs, resorcinole is not, but pyrogallol is.

Just curious: based on Kendall-Pelz rule, would 1,3-Butadiene-1,4-diamine also work as a developer? I know, a substance which has a flash point below 65°C falls squarely into the "things I won't work with" category, I ask this from a purely theoretical stand point.

For the basic discovery of resonance see "Kekule"

http://en.wikipedia.org/wiki/Friedrich_August_Kekulé_von_Stradonitz

For advanced work see "Hammett".

http://en.wikipedia.org/wiki/Hammett_equation

For the answer as to whether 1,3-Budadiene-1,4-diamine is a developer consider the exceptions to this rule and consider the extensions by Umberger.

I would not work with it either and I doubt if anyone would in a small darkroom, and therefore there will never be any real way to prove whether it is or is not.

PE

Rudeofus said:

It appears that the A-(C=C)n-A' notation is a strongly simplified version of Kendall and Pelzs original research work, and just looking at A-(C=C)n-A' does not tell me imbecile, why hydroquinone and catchechol are devs, resorcinole is not, but pyrogallol is.

Click to expand...

The notation is directly from Mason although changed due to the limitations of the APUG text editor.

If you look at the rule it says that the chain linking the two developer groups must begin and end with a double bond when n > 0.. This is true for catechol and hydroquinone but not for resorinol where it ends with a single bond.

Photo Engineer said:

I would not work with it either and I doubt if anyone would in a small darkroom, and therefore there will never be any real way to prove whether it is or is not.
PE

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Ron, I once synthesised pyridylhydroquinone from quinone and pyridine as mentioned in Glafkides. IIRC, it is a fine grain developing agent similar in action to chlorhydroquinone. The compound consisted of fine yellow needles. When used in a developer the solution turned dark red when the alkali was added. The one drawback was that it slowly decomposed in solution to produce pyridine which has a VERY bad smell. For those who are unfamiliar with this chemical, once you get a whiff of pyridine it seems to stay in your nose for quite some time. Ethyl ether has the same effect. No one would want that smell in their darkroom.

Just a caution -- do not attempt anything like this at home. Quinone is a highly reactive and very nasty chemical. The LD50 dose is 25 mg/kg for mice.

Gerald C Koch said:

If you look at the rule it says that the chain linking the two developer groups must begin and end with a double bond.

Click to expand...

That clears it up quite a bit. Since vinyl alcohol changes its structure to acetaldehyde as soon as it gets a chance to, I naively assume that this would make the linear structure -C=C-OH highly unstable in general, while OH attached to a benzene ring is stable hence has a chance to work as dev before bad things happen to it. There is no reason to risk blow up our little dark rooms with alkadiene-diols after all.

It may all be obvious to real chemists, but for extra clarity we could rephrase the A-(C=C)n-A' rule as A-C=C-whatever-C=C-A', maybe with the hint that the C=C double bonds may(and in all likeliyhood will) come from an aromatic ring.

The double bonds do not have to be in an aromatic ring. The rule also fits aliphatic compounds such as the butadiene compound mentioned above. Remember cases like hydrazine and hydroxyamine where n = 0.

Jerry;

Pyridine is also detrimental to having children. It causes loss in fertility and mutations.

In any event, I am just pointing out that there are a lot of potential developers out there, and a lot of them do not fit a general rule. The more accurate rules are based on redox potential vs AgX and the ability to discriminate between unexposed silver halides and exposed silver halides.

PE

I love the smell of diethyl ether! One of the better smelling solvents I work with.

Pyridine, some people really don't like it, but after using for 20 years as part of the color reagent for the cyanide test, I'm used to it.

Ketones (other than acetone) are another one I don't like, and ethyl acetate is not so nice either.