The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
$$R-Cl+KOH\text{(aq)} \rightarrow R-OH+KCl$$
The ionization of aqueous KOH produces hydroxide ions which are strong nucleophiles. Hence, alkyl chlorides undergo substitution to form alcohol.
$$R-CH_2-CH_2-Cl+KOH\text{(alc)}\rightarrow R-CH=CH_2+KCl+H_2O$$
Alcoholic KOH solution gives alkoxide ion which is a strong base. It abstracts $$\displaystyle \beta$$ hydrogen atom of alkyl chloride. A molecule of HCl is eliminated and an alkene is formed.
Note: The basicity of hydroxide ion is much lower than the basicity of alkoxide ion as hydroxide ion is significantly hydrated in aqueous solution.
Hence, hydroxide ion cannot abstract $$\displaystyle \beta$$ hydrogen atom of alkyl chloride.
Consider the following reactions: (a) $$(CH_3)_3CCH(OH)CH_3\xrightarrow{conc.H_2SO_4}$$ (b) $$(CH_3)_2CHCH(Br)CH_3\xrightarrow{alc.KOH}$$ (c) $$(CH_3)_2CHCH(Br)CH_3\xrightarrow{(CH_3)_3CO^{\ominus}K^{\oplus}}$$ (d) $$(CH_3)_2\underset{\,\,\,\,OH}{\underset{|}{C}}-CH_2-CHO\xrightarrow{\Delta}$$ Which of these reaction(s) will not produce Saytzeff product?
Which statement is correct about the reactions?
Refer to Q. No. 14 and find out which statement is correct.
Which statement is correct about the following reaction?