Q12P-d - Organic Chemistry (Mcmurry)

Question

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

$-C{H}_{2}C{H}_3,-\mathrm{C}\equiv CH,-\mathrm{C}\equiv \mathrm{N},-C{H}_{2}OC{H}_3$

Step-by-Step Solution

Answer

Higher to lower ranking:$-{\mathrm{CH}}_2{\mathrm{OCH}}_3>-\mathrm{C}\equiv \mathrm{N}>-\mathrm{C}\equiv CH>-{\mathrm{CH}}_2{\mathrm{CH}}_3$

Step 1: E and Z designation

Disubstituted alkenes with two substituents other than hydrogen on the double bond are only named using the cis-trans system. But when trisubstituted and tetra substitution 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 $-{\mathrm{CH}}_2{\mathrm{OCH}}_3$ carbon is attached two hydrogens and one oxygen.
• In $-\mathrm{C}\equiv \mathrm{N}$carbon is attached to three nitrogen.
• In carbon $-\mathrm{C}\equiv CH$ is attached to three carbon.
• In carbon $-{\mathrm{CH}}_2{\mathrm{CH}}_3$ is attached to two hydrogens and one carbon.

So, higher to lower-ranking: $-{\mathrm{CH}}_2{\mathrm{OCH}}_3>-\mathrm{C}\equiv \mathrm{N}>-\mathrm{C}\equiv CH>-{\mathrm{CH}}_2{\mathrm{CH}}_3$