Question

$\underset{6}{C}{H}_3-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}{H}_2-\underset{2}{C}\equiv \underset{1}{C}H$
The hybridizations of the carbons numbered 1, 3 and 5 in the given hydrocarbon are respectively:

• A. $s{p}^3,s{p}^2,sp$
• B. $s{p}^2,sp,s{p}^3$
• C. $sp,s{p}^3,s{p}^2$
• D. $sp,s{p}^2,s{p}^3$

Answer 1

The correct option is C $sp,s{p}^3,s{p}^2$

$-C-C-⟹s{p}^3-C=C-⟹s{p}^2-C\equiv C-⟹sp=C=C=⟹sp$

$\underset{6}{C}{H}_3-\underset{5}{C}H=\underset{4}{C}H-\underset{3}{C}{H}_2-\underset{2}{C}\equiv \underset{1}{C}H$

$\stackrel{s{p}^3}{C}{H}_3-\stackrel{s{p}^2}{C}H=\stackrel{s{p}^2}{C}H-\stackrel{s{p}^3}{C}{H}_2-\stackrel{sp}{C}\equiv \stackrel{sp}{C}H$