1. aqueous koh gives alcohols but alcofolic koh gives alkene on reaction with alkyl halide . why?
2.para chlorobenzene ha shighr solubility compared to ortho and meta isomers . why?
1. aqueous koh gives alcohols but alcofolic koh gives alkene on reaction with alkyl halide . why?
2.para chlorobenzene ha shighr solubility compared to ortho and meta isomers . why?
1. Alkyl halides in the presence of aqueous KOH, under goes substitution reaction to give alcohols.
E.g. Alkyl chloride with aqueous KOH gives substitution product alcohol. Here OH- acts as a strong nucleophile facilitates the substitution reaction and replaces halogen from alkyl halide with OH- ion.
RCl + KOH (aq.) → ROH
Alkyl halides in the presence of alcoholic KOH under goes elimination reaction to give alkenes. Alkyl bromide containing ß-hydrogen results ß- elimination ( or 1,2-elimination) to give the product alkene.
E.g. When ethyl bromide treated with alc. KOH, the C2H5O- being a stronger base than OH- ion facilitates elimination reaction abstracts the ß-hydrogen of ethyl bromide to give alkene.
CH3CH2Br + KOH (alc.) → H2C=CH2 + KBr + H2O
2. para-dichloro benzene has lowest solubility compared to meta and ortho-isomers.
p-dichlorobenzene is symmetrical and best fits into a crystal lattice than meta and ortho-isomers. Therefore p-dichlorobenzene has strong intermolecular forces of attraction as compared to the ortho and meta isomers.
Hence during dissolution large amount of energy is needed to dissolve the para isomer. Hence the para isomer is less soluble than ortho and meta isomer.
1. Alkyl halides in the presence of aqueous KOH, under goes substitution reaction to give alcohols.
E.g. Alkyl chloride with aqueous KOH gives substitution product alcohol. Here OH- acts as a strong nucleophile facilitates the substitution reaction and replaces halogen from alkyl halide with OH- ion.
RCl + KOH (aq.) → ROH
Alkyl halides in the presence of alcoholic KOH under goes elimination reaction to give alkenes. Alkyl bromide containing ß-hydrogen results ß- elimination ( or 1,2-elimination) to give the product alkene.
E.g. When ethyl bromide treated with alc. KOH, the C2H5O- being a stronger base than OH- ion facilitates elimination reaction abstracts the ß-hydrogen of ethyl bromide to give alkene.
CH3CH2Br + KOH (alc.) → H2C=CH2 + KBr + H2O
2. para-dichloro benzene has lowest solubility compared to meta and ortho-isomers.
p-dichlorobenzene is symmetrical and best fits into a crystal lattice than meta and ortho-isomers. Therefore p-dichlorobenzene has strong intermolecular forces of attraction as compared to the ortho and meta isomers.
Hence during dissolution large amount of energy is needed to dissolve the para isomer. Hence the para isomer is less soluble than ortho and meta isomer.
