Question

Consider the following reactions.
$\left(1\right)\left(C{H}_3{)}_{3}CCH\right(OH)C{H}_3\underset{\Delta }{\overset{\text{conc.}{H}_{2}S{O}_4}{\to }}$
$\left(2\right)\left(C{H}_3{)}_{2}C{H}_{2}C{H}_{2}C{H}_2\right(Br)\stackrel{\text{alc. KOH}}{\to }$
$\left(3\right)\left(C{H}_3{)}_{2}CHCH\right(Br)C{H}_3\stackrel{(C{H}_3{)}_{3}C{O}^{–}{K}^{+}}{\to }$
$\left(4\right)(C{H}_3{)}_2\underset{OH}{\underset{|}{C}}-C{H}_2-CHO\stackrel{\Delta }{\to }$
Which of these reactions(s) will produce Saytzeff product as major product?

• A. (4) only
• B. (1), (2) and (4)
• C. (3) only
• D. (2) and (4)

Answer 1

The correct option is B (1), (2) and (4)

$\left(1\right)\left(C{H}_3{)}_{3}CCH\right(OH)C{H}_3\underset{\Delta }{\overset{\text{conc.}{H}_{2}S{O}_4}{\to }}\underset{\text{More stable alkene}}{(C{H}_3{)}_{3}C-CH=C{H}_2}$
$\left(2\right)\left(C{H}_3{)}_{2}CHCH\right(Br)\stackrel{\text{alc. KOH}}{\to }\underset{\text{More stable alkene}}{(C{H}_3{)}_{2}C=CH-C{H}_3}$
$\left(3\right)\left(C{H}_3{)}_{2}CHCH\right(Br)C{H}_3\stackrel{(C{H}_3{)}_{3}C{O}^{–}{K}^{+}}{\to }\underset{\text{less stable alkene}}{(C{H}_3{)}_{2}HCCH=C{H}_2}$
$\left(4\right)(C{H}_3{)}_2\underset{OH}{C\underset{|}{}}-C{H}_2-CHO\stackrel{\Delta }{\to }\underset{\text{More stable alkene}}{(C{H}_3{)}_{2}C=CH-CHO}$

Bulkier base like $(C{H}_3{)}_3{O}^{–}{K}^{+}$ always perferes Hoffmann product (less stable alkene).
In case of 1, 2 and 4, elimination products are according to Saytzeff Rule (more stable alkene).