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Ethyl alcohol is heated with conc H2SO4 the product formed is
Which of the following is basic
The compound which reacts fastest with Lucas reagent atroom temperature is (1981 - 1 Mark)
The order of reactivity of alcohol with Lucas reagent is tert. > sec. > pri.
Lucas test is based on the difference in the three types of alcohols (having 6 or less carbon) towads Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride) at room temperature.
The tertiary alcohols produce turbidity immediately, the secondary alcohols give turbidity within 5 – 10 minutes, and the primary alcohols do not give turbidity at all, at room temperature. Hence 2-methylpropan-2-ol (a 3° alcohol) reacts fastest.
A compound that gives a positive iodoform test is(1982 - 1 Mark)
Compounds having groups show positive iodoform test.
(pentanone-2) gives this test.
Diethyl ether on heating with conc. HI gives two moles of (1983 - 1 Mark)
An industrial method of preparation of methanol is : (1984 - 1 Mark)
When phenol is treated with excess bromine water, it gives: (1984 - 1 Mark)
The –OH group in phenol, being activating group, facilitates substitution in the o- and p-positions.
HBr reacts fastest with : (1986 - 1 Mark)
Reactions involving cleavage of carbon-oxygen bond, (C – OH) follows the following order :
Tertiary > Secondary > Primary
Which of the following compounds is oxidised to preparemethyl ethyl ketone? (1987 - 1 Mark)
Secondary alcohols oxidise to produce kenone
Phenol reacts with bromine in carbon disulphide at lowtemperature to give (1988 - 1 Mark)
NOTE : In absence of CS2, polyhalogenation in o- and p-positions takes place.
Chlorination of toluene in the presence of light and heatfollowed by treatment with aqueous NaOH gives(1990 - 1 Mark)
When phenol is reacted with CHCl3 and NaOH followed by acidification, salicyladehyde is obtained. Which of the
following species are involved in the above mentioned reaction as intermediate?
Riemer-Tiemann reaction involves electrophilic
substitution on the highly reactive phenoxide ring.
The compound that will react most readily with NaOH toform methanol is (2001S)
Compound (CH3)4N+I– is most reactive due to (i) better
leaving group, I– and (ii) due to the fact that the methyl group, with positive N, is more electron deficient.
Hence this group is more reactive towards nucleophile, OH–
1-Propanol and 2-propanol can be best distinguished by (2001S)
Fehling solution is a weak oxidising agent which can oxidise aldehyde but not ketone.
Primary alcohols undergoes oxidation with alkaline KMnO4, acidic dichromate and conc. H2SO4 to give acids, whereas with Cu they give aldehydes.
NOTE : This reaction is an example of Williamson's synthesis.
C2H5O– will abstract proton from phenol converting the latter into phenoxide ion. This would then make nucleophilic attack on the methylene carbon of alkyl iodide forming C6H5OC2H5. But if C2H5O– is in excess,
C2H5OC2H5 will be formed.C2H5O– is a better
nucleophile than C6H5O– (phenoxide) ion because in the former the negative charge is localised over oxygen, while in the latter it is delocalised over the whole molecular framework. So, it is C2H5O– ion that would
make nucleophilic attack at ethyl iodide to give diethyl ether (Williamson’s synthesis).
The product of acid catalyzed hydration of 2-phenylpropeneis (2004S)
NOTE : Addition of water to 2-phenylpropene follows Markownikov’s rule.
The best method to prepare cyclohexene from cyclohexanolis by using (2005S)
Conc. HCl, HBr and conc. HCl + ZnCl2 all are
nucleophiles, thus convert alcohols to alkyl halides.
However, conc. H3PO4 is a good dehydrating agent which converts an alcohol to an alkene.
The increasing order of boiling points of the below mentioned alcohols is (2006 - 3M, –1)
Among the given compounds, hydroxybenzene (IV) has least molar mass and therefore possess least boiling point. Among the three isomeric dihydroxybenzenes, 1,2-dihydroxybenzene (I) forms intramolecular
H-bonding with the result it will not form intermolecular H-bonding leading to lowest boiling point. On the other hand 1,3-dihydroxybenzene (II) and 1, 4-dihydroxybenzene (III) do not undergo chelation, hence they will involve extensive intermolecular H-bonding leading to higher boiling point. Further intermolecular hydrogen bonding is stronger in the p-isomer than the m-isomer hence former has highest b.p. Thus the decreasing order of boiling points is
III > II > I > IV.
In the reaction the products are
For the identification of b-naphthol using dye test, it isnecessary to use (JEE Adv. 2014)
In dye test, phenolic — OH group is converted to — O– which activates the ring towards electrophilic aromatic substitution