Practice free Alcohol, phenol & ether (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
In the following Sequence of reactions,$ CH_3 CH_2 OH \xrightarrow [ ] { {P + I_2 }} A \xrightarrow [ { ether} ] { { Mg }} B \xrightarrow [ ] { {HCHO} } C \xrightarrow [ ] { {H_2 O } }D $ , the compound D is:
Acid catalysed hydration of alkenes except ethene leads to the formation of
During dehydration of alcohol to alkenes by heating with Conc $H_2SO_4$ , the initiation step is:
In the dehydration of alcohols to alkenes using concentrated $H_2SO_4$, the initiation step is the protonation of the alcohol molecule. This step involves the alcohol accepting a proton from the $H_2SO_4$, forming an oxonium ion, which makes the -OH group a better leaving group.
which of the following compounds will give positive iodoform test ?(I) 3- methyl propan-2-ol (II) I - methyl cyclopentanal (III ) I - phenyl propan-1-ol (Iv) 3- phenyl propan-2-ol.
The iodoform test is used to identify compounds with the structure $CH_3C(OH)R$ or $CH_3COR$. Compound (I) 3-methyl propan-2-ol and compound (IV) 3-phenyl propan-2-ol both have the $CH_3C(OH)R$ group, and will therefore give a positive iodoform test.
Propan -l-ol and propan-2-ol can be distinguished by
$ CH_3CH_2OH $ cannot be prepare by which of the following reaction ?
Ethanol ($CH_3CH_2OH$) cannot be prepared by the reaction of ethyl chloride with alcoholic potassium hydroxide. This reaction typically results in the formation of ethene through an elimination reaction (dehydrohalogenation), not ethanol.
The most suitable reagent for the conversion of primary alcohol into aldehyde with the same number of carban is
Pyridinium chlorochromate (PCC) is a mild oxidizing agent that converts primary alcohols into aldehydes without further oxidation to carboxylic acids, which makes it the most suitable reagent for this conversion.
An organic compound "X" on treatment with PDC in $CH_2Cl_2 $ gives compound "Y". Compound "Y", reacts with $ I_2 $ and alkali to form yellow precipitate. The compound "X" is
The compound 'X' is ethanol. When ethanol is treated with Pyridinium Dichromate (PDC) in $CH_2Cl_2$, it gets oxidized to ethanal (compound 'Y'). Ethanal reacts with $I_2$ and alkali to form iodoform (yellow precipitate), confirming the presence of a methyl ketone or an aldehyde with a methyl group adjacent to the carbonyl group.
The correct order of boiling points is for n-Butyl alcohol tert - Butyl alcohol (I) (II) iso - Butyl alcohol Sec- Butyl alcohol (I) (IV)
The boiling point of glycerol is more than propanol because of
Glycerol has a higher boiling point than propanol primarily because of hydrogen bonding. Glycerol has three hydroxyl (OH) groups capable of forming extensive hydrogen bonds, leading to a higher boiling point compared to propanol, which has only one hydroxyl group.
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