The synthesis of alkyl fluorides is best accomplished by
(JEE Main 2015)
Alkyl fluorides can be prepared by action of mercurous fluoride or antimony trifluorides (inorganic fluorides) on corresponding alkyl halide.
But, when action of Nal/acetone takes place on alkyl chloride bromide, alkyl iodide forms. This reaction is called 'Finkelstein reaction'
Free radical fluorination is highly explosive reaction. So not preferred for the preparation of fluoride.
The major organic compound formed by the reaction of 1,1,1 -trichloroethane with silver powder is
Compound A, (C8H9Br) gives a white precipitate when warm ed with alcoholic AgNO3. Oxidation of A gives an acid B, (C8H6O4). B easily forms anyhydride on heating. Identify the com pound A .
(JEE Main 2013)
Com pound A gives a precip itate w ith alcoholic AgNO3 (here white is misprinting because the colour of ppt is light yellow), so it must contain Br in side chain.
On oxidation, it gives C8H6O4, which shows the presence of two alkyl chains attached directly with the benzene nucleus.
Since compound B gives anhydride on heating, the two alkyl substituent must occupy adjacent (1,2) position.
Thus, A must b
and the reactions are as follows
A solution of (-1)-chloro-1 -phenylethane in toluene racemises slowly in the presence of a small amount of SbCI5, due to the formation of
(JEE Main 2013)
The given compound, (-1)- chloro-1-phenylethane in the presence of SbCI5 forms a carbocation.
Since, the carbocation is a planar species, therefore it can be attacked by SbCl6- either from the front or back side of the carbocation with equal ease. As a result, 50:50 mixture of two enantiomers of 1 -chloro-1 -phenylethane undergoes racemisation due to the formation of a carbocation intermediate.
How many chiral compounds are possible on monochlorination of 2-methyl butane?
2 -methylbutane on monochlorination gives 4 isomers, among which I and III are chirai in nature.
Hence, 2 chiral compounds are formed in the above reaction.
Iodoform can be prepared from all except
Iodoform reaction is given by alcohols and ketones containing group, respectively. Thus, among the given compounds, isobuty alcohol does not contain
Hence, it does not give iodoform reaction on treatment with I2 / NaOH.
Hence, compounds (a), (b) and (c) will give iodoform while compound (d) (isobutyl alcohol) does not give any iodoform reaction
Consider the following bromides,
The reactivity of SN1 reaction depends upon the stability of the intermediate, carbocation formed during these reactions. The stabililty order of the carbocation formed from the given species is
Hence, the reactivity order of the given bromide towardsSN1 reaction is
The organic chloro compound, which shows complete stereochemical inversion during an SN2 reaction is
Nucleophilic substitution bimolecular (SN2) prefers less sterically hindered site to attack. Lesser the steric hindrance faster the SN2 reaction. So ease of reaction is 1° > 2° > 3°.
SN2 involves inversion of configuration stereo, chemically. Since, 1° alkyl halides are prone to SN2 reactions, therefore CH3CI undergoes complete strerochemical inversion.
Which of the following reactions will yield 2, 2-dibromopropane?
The decreasing order of the rate of the above reaction with nucleophile (Nu-) A and D is [Nu = (A) PhO- , (B) AcO- , (C ) HO- , (D) CH3O- ]
Nucleophilicity order is
Alkyl halides react with dialkyl copper reagents to give
2-methyl butane on reacting with bromine in the presence of sunlight gives mainly
Elimination of bromine from 2-bromobutane results in the formation of
Stability of (I) > (II), hence (I) is predominant.
Reaction of one molecule of HBr with one molecule of 1 , 3-butadiene at 40°C gives predominantly
Addition is through the formation of allylic carbocation.
Under mild conditions (temperature = - 80°C) kinetic product is the 1, 2-addition product and under vigorous conditions, (temp. = 4°C) thermodynamic product is the 1 , 4-addition product.
Thus, 1-bromo-2-butene is the major product under given condition.