In the hydrocarbonCH26−CH5=CH4−CH23−C2≡CH1The state of hybridization of carbons 1, 3 and 5 are in the following sequence

A
$sp, {sp}^{2}, {sp}^{3}$
B
${sp}^{3}, {sp}^{2}, sp$
C
${sp}^{2}, sp, {sp}^{3}$
D
$sp, {sp}^{3}, {sp}^{2}$

Solution:

${CH}_{3} - \underset{\underset{{sp}^{2}}{↓}}{\overset{5}{CH}} = CH - \underset{\underset{{sp}^{3}}{↓}}{\overset{3}{{CH}_{2}}} - C = \overset{1}{\underset{\underset{sp}{↓}}{CH}}$

In the hydrocarbon $\underset{6}{{CH}_{2}} - \underset{5}{CH} = \underset{4}{CH} - \underset{3}{{CH}_{2}} - \underset{2}{C} \equiv \underset{1}{CH}$
The state of hybridization of carbons $1, 3$ and $5$ are in the following sequence

Solution:

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