Q12P-b - Organic Chemistry (Mcmurry)

Question

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

$-{\mathrm{CH}}_3,-{\mathrm{CH}}_2{\mathrm{CH}}_3,-\mathrm{CH}={\mathrm{CH}}_2,{\mathrm{CH}}_2\mathrm{OH}$

Step-by-Step Solution

Answer

Higher to lower-ranking:$-{\mathrm{CH}}_2\mathrm{OH}>-\mathrm{CH}={\mathrm{CH}}_2>-{\mathrm{CH}}_2{\mathrm{CH}}_3>-{\mathrm{CH}}_3$

Step 1: E and Z designation

Cis-trans naming is limited to alkenes with two substituents other than hydrogen on the double bond. 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 $-{\mathrm{CH}}_2\mathrm{OH}$ carbon is attached to two hydrogens and one oxygen.
• In $-\mathrm{CH}={\mathrm{CH}}_2$ carbon is attached to two carbon and one hydrogen.
• In $-{\mathrm{CH}}_2{\mathrm{CH}}_3$carbon is attached to two hydrogens and one carbon.
• In $-{\mathrm{CH}}_3$carbon is attached to three hydrogens only.
• So, higher to lower-ranking:$-{\mathrm{CH}}_2\mathrm{OH}>-\mathrm{CH}={\mathrm{CH}}_2>-{\mathrm{CH}}_2{\mathrm{CH}}_3>-{\mathrm{CH}}_3$