Question

Give reason for the following:
i. Ethyl iodide undergoes $S{{N}^{2}}$faster than ethyl bromide.
ii. $\left( \pm \right)2-$ Butanol is optically inactive.
iii. $C-X$bond length in halobenzene is smaller than $C-X$ bond length in $C{{H}_{3}}-X$.

Answer 1

Reactivity of iodine is higher and breaking of connection among ethyl and iodine is simpler when contrasted with bromine. Butan $-2-ol$ has asymmetric carbon (chiral center) with four different groups attached to the central carbon.

Complete step by step answer:
Iodide is a superior leaving bunch in light of its bigger size than bromide, consequently, ethyl iodide goes through $S{{N}^{2}}$ response quicker than ethyl bromide. Because in ethyl iodide, iodide(I) is acting as a best leaving group among all the halide ions so the rate of SN​2 reaction is directly proportional to leaving group ability.
Basicity decreases with increase in size
Iodine is a better leaving group
(ii) $\left( \pm \right)2-$ butanol is a racemic combination. It is a blend which contains two enantiomers in equivalent extent and accordingly, have zero optical revolution because of inner remuneration. Accordingly, it is optically idle. It is a racemic mixture which is optical also because of external compensation.
(iii) In halobenzenes (like chlorobenzene), the solitary pair of electrons on halogen iota is delocalised on the benzene ring. Therefore, $C-X$bond ($C-Cl$ ) bond in the event of chlorobenzene) procures some twofold bond character while in $C{{H}_{3}}-X$, $C-X$ bond is an unadulterated single security. Along these lines, $C-X$ bond in halobenzene is more limited than in $C{{H}_{3}}-X$.

Note:
Enantiomers have indistinguishable compound and physical properties, aside from their impact on plane energized light and response with other chiral atoms. Due to resonance in halobenzene it has less bond length value as compared to $C{{H}_{3}}-X$ .