Q12P-c - Organic Chemistry (Mcmurry)

Question

Rank the substituents in each of the following sets according to the sequence rules:

$-{\mathrm{CO}}_2\mathrm{H},-{\mathrm{CH}}_2\mathrm{OH},-\mathrm{C}\equiv \mathrm{N},-{\mathrm{CH}}_2{\mathrm{NH}}_2$

Step-by-Step Solution

Answer

Higher to lower-ranking: 

Step 1: E and Z designation

Only disubstituted alkenes with two substituents other than hydrogen on the double bond can use the cis-trans system of naming. But when trisubstituted and tetrasubstituted of the double bond is present then the E and Z system is used for describing alkene stereochemistry.

Step 2: Rule for ranking substituents

• Consider each of the double-bond carbons separately and ranks the two substituents according to the atomic number of the first atom in each. An atom with a higher atomic number will rank higher.

 

• If the first atom same then, look at the second, third, or fourth atoms away from the double-bond until the first difference is found.

 

• Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.

Step 3: Ranking of given groups

As ranking is atomic number dependent, more the atomic number higher will be ranking. So, 

• In $-{\mathrm{CO}}_2\mathrm{H}$ carbon is attached to three oxygen.
• In $-{\mathrm{CH}}_2\mathrm{OH}$ carbon is attached to two hydrogens and one oxygen.
• In $-\mathrm{C}\equiv \mathrm{N}$ carbon is attached to three nitrogen.
• In $-{\mathrm{CH}}_2{\mathrm{NH}}_2$carbon is attached to two hydrogens and one nitrogen.

So, higher to lower-ranking:$-{\mathrm{CO}}_2\mathrm{H}>-{\mathrm{CH}}_2\mathrm{OH}>-\mathrm{C}\equiv \mathrm{N}>-{\mathrm{CH}}_2{\mathrm{NH}}_2$