Which one is formed when sodium phenoxide is heated with ethyl iodide ? 
Lucas reagent is 
Lucas reagent is conc. HCl + anhyd. ZnCl2.
The compound which reacts fastest with Lucas reagent at room temperature is 
The rates of reaction with lucas reagen t follows the order. 3° alcohol > 2° alcohol > 1° alcohol since carbocations are formed as intermediate, more stable the carbocation, higher will be the reactivity of the parent compound (alcohol). 2-Methylpropan-2-ol generates a 3º carbocation, so it will react fastest; other three generates either 1º or 2º carbocations.
Which one of the following compounds will be most readily attacked by an electrophile ?
Due to strong electron-donating effect of the OH group, the electron density in phenol is much higher than that in toluene, benzene and chlorobenzene and hence phenol is readily attacked by the electrophile.
Propene, CH3CH = CH2 can be converted into 1-propanol by oxidation. Indicate which set of reagents amongst the following is ideal to effect the above conversion ? 
KMnO4 (alkaline) and OsO4/CH2Cl2 are used for hydroxylation of double bond while O3/Zn is used for ozonolysis. Therefore, the right option is (c), i.e.,
When phenol is heated with CHCl3 and alcoholic KOH when salicyladehyde is produced. This reaction is known as [1988, 89]
How many isomers of C5H11 OH will be primary alcohols ? 
Four primary alcohols of C5H11OH are possible. These are:
Methanol is industrially prepared by 
Water gas is mixed with half of its volume of hydrogen.The mixture is compressed to approximately 200 – 300 atmospheres. It is then passed over a catalyst |ZnO + Cr2O3| at 300°C. Methyl alcohol vapours are formed which are condensed
HBr reacts fastest with
Greater the stability of the intermediate carbocation, more reactive is the alcohol.
Since 2-methylpropan-2-ol generates 3° carbocation, therefore, it reacts fastest with HBr.
Wh en phenol is tr eated with excess bromine water. It gives 
With Br2 water, phenol gives 2, 4, 6-tribromophenol.
Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of 
Due to H-bonding, the boiling point of ethanol is much higher than that of the isomeric diethyl ether.
Increasing order of acid strength among p-methoxyphenol, p-methylphenol and p-nitrophenol is 
Electron-donating groups (– OCH3, – CH3 etc.) tend to decrease and electron withdrawing groups (– NO2, – OCH3 etc.) tend to increase the acidic character of phenols. Since – OCH3 is a more powerful electron-donating group than – CH3 group, therefore, p-methylphenol is slightly more acidic than p-methoxyphenol while pnitrophenol is the strongest acid. Thus, option (d), i.e. p-methoxyphenol, pmethylphenol, p-nitrophenol is correct.
Which one of the following on oxidation gives a ketone ? 
Secon dary alcoh ols on oxidation give ketones.
Note : – Primary alcohols from aldehydes.
What is formed when a primary alcohol undergoes catalytic dehydrogenation ? 
1° Alcohols on catalytic dehydrogenation give aldehydes.
The stablest among the following is 
Due to –I-effect of the three C–Cl-bonding between CI and C-atom of the OH group, CCl3 CH (OH)2 is most stable.
1-Phenylethanol can be prepared by the reaction of benzaldehyde with 
Reaction of with RMgX leads toformation of
We know that
The ionization constant of phenol is higher than that of ethanol because : 
The acidic nature of phenol is due to the formation of stable phenoxide ion in solution
The phenoxide ion is stable due to resonance.
The negative charge is delocalized in the benzene ring which is a stabilizing factor in the phenoxide ion and increase acidity of phenol. wheras no resonance is possible in alkoxide ions (RO–)derived from alcohol. The negative charge is localized on oxygen atom.
Thus, alcohols are not acidic.
Propan - 1- ol may be prepared by the reaction of propene with 
Hence C is correct.
Which of the following is correct ? 
n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?
Primary alcohol on oxidation give aldehyde which on further oxidation give carboxylic acid whereas secondary alcohols give ketone.
When phenol is treated with CHCl3 and NaOH, the product formed is 
Which of the following will not form a yellow precipitate on heating with an alkaline solution of iodine? 
CH3OH does not have -CH(OH)CH3 grouping hence it will not form yellow precipitate with an alkaline solution of iodine (haloform reaction).
The major organic product in the reaction, CH3 — O — CH(CH3)2 + HI → Product is 
First HI will break into H+ and I-. Then H+ attacks on ether to form CH3O+HCH(CH3)2
Being an excellent nucleophile, I- attacks where it is easy to attack and so it will go for less bulky CH3 to form CH3I and (CH3)2CHOH.
Ethylene oxide whent reated with Grignard reagent yields 
Ethylene oxide when treated with Grignard Reagent gives primary alcohol.
Which of the following compounds will form hydrocarbon on reaction with Grignard reagent ?
Consider the following reaction, the product Z is:
H2COH · CH2OH on heating with periodic acid gives: 
1, 2 – Diols, when treated with an aqueous solution of periodic acid give aldehyde or ketones
Note that a 1° alcohol gives CH2O. Since in glycol both the OH groups, are primary hence give 2 molecules of CH2O as by product.
Consider the following reaction: 
The product Z is:
Which one of the following compounds has the most acidic nature? 
The presence of electron withdrawing substituent increases the acidity while electron releasing substituent, when present, decrease the acidity.
Phenyl is an electron withdrawing substituent while -CH3 is an electron releasing substituent, Moreover, phenoxide ion is more resonance stabilised as compared to benzoyl oxide ion, thus, release proton more easily. That's why is a strong acid among the given.
Among the following four compounds (i) phenol (ii) methylphenol (iii) meta-nitrophenol (iv) para-nitrophenol the acidity order is :
(– I and – M effects, (only – I effect) both increase acidity)
(+ I effect of CH3 group decreases acidity)
∴ Correct choice : (b)
When glycerol is treated with excess of HI, it produces: 
Glycerol when treated with excess HI produces 2–iodopropane
∴ Correct choice : (b)
In the following sequence of reactions the end product (C) is :
Which of the following compounds can be used as antifreeze in automobile radiators ? [2012 M]
Glycol is used as an antifreeze in automobiles.
Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
The reaction will proceed via SN1 or SN2 based on nature of alkyl group. If alkyl group attached is 3°. The reaction will proceed through the SN1 mechanism and if alkyl group is primary reaction will proceed through SN2 mechanism.
Phenol is distilled with Zn dust followed by Friedel Crafts alkylation with propyl chloride in the presence of AlCl3 to give a compound (B). (B) is oxidised in the presence of air to form the compound (C). The structural formula of (C) is [NEET Kar. 2013]