Title 10.HALOALKANES AND HALOARENES - Explanations.pdf ✅

1 (a) In SN2 reaction, nucleophile and alkyl halide react in one step. Thus, tertiary carbon is under steric hindrance thus reaction does not take place until (C-Br) bond breaks Which is the SN1reaction. 2 (b) CH3 CH3 | | CH3 − C − Br + CH3ONa ⟶ CH3 − C = | CH3 CH2 + CH3OH + NaBr t-butyl bromide sodium 2-methyl propene methanol methoxide (isobutylene) 3 (d) Iodoform test is given by compounds which have CH3CO or CH3CHOH group. H | (a)H3C − C − CH3 (b)CH3CH2OH | ethyl alcohol OH Iso-propyl alcohol O ∥ (c) CH3 − C − H ethanal (i) iso-propyl alcohol, ethanol ad ethanal all have CH3CO or CHOH group, therefore they give iodoform test. (ii) Benzyl alcohol does not have CH3CO − or CHOH group, | CH3 Therefore, it does not give iodoform test. 4 (a) C2H5Cl + KCN C2 H5OH → C2H5CN + KCl (X) C2H5CN H3O+,2H2O → ∆ C2H5COOH + NH3 (Y)or (C3H6O2) So, the molecular formula of the Y is C3H6O2. 5 (d) Chloride is an 1° aliphatic carbon which is substituted easier in comparison to iodide which is arylic and more stable due to delocalisation hence, difficult to substitute. 7 (d) Solvolysis of haloalkanes follows first order kinetics. During this process an intermediate carbocation is formed. Therefore, the halohydrocarbon which gives more stable carbocation undergoes solvolysis readily. 11 (c) γ-isomer of cyclohexane hexachloride is strong pesticide. It is also known as lindane. 12 (a) CHCl3 + 3C2H5ONa → CH(OC2H5 )3 + 3NaCl Ethy ortho formate 13 (a) ROH + SOCl2 → RCl + SO2 ↑ + HCl ↑ ∵ SO2 and HCl are gaseous by-products and can be removed easily to get pure alkyl halide. ∴ It is best method for preparation of alkl halide. 14 (c) Iodoform test is given by only those compounds which conatain either
CH3C = O or CH3CH − OH group | | (a) CH3CH − OH (b) CH3CH − OH | | H CH3 ethyl alcohol propanol-2 (c) CH3CH2CH2OH (c) CH3 − C = O Propanol-1 | H ethanal Hence, propanal-1 due to absence of above given groups, does not give positive iodoform test. 15 (a) CH3NH2 CH3I → −HI (CH3 )2NH CH3I → −HI (CH3 )3N CH3I → (CH3 )4N +I − Hence, three molecules of CH3I is used. 17 (a) Benzyl chloride is more reactive than alkyl halides. Benzyl carbocation is stabilised by resonance hence, benzyl chloride easily gives nucleophilic substitution reaction. 18 (c) Iodoform test is given by those compounds which has CH3 − C − C/H or CH3 − CH − units. ∥ | O OH Hence, this test is not given by phenol (C6H5 − OH). 19 (c) An alkyl halide on heating with dry silver oxide gives ether. 2R − X + Ag2O ∆ → R − O − R + 2AgX alkyl halide dry ether 20 (a) RCl and RBr can be prepared by free radical halogenation of alkanes while RF and RI cannot be prepared. With F2, the reaction is not only explosive but also brings cleavage of C-C bonds while with I2 the reaction is too slow to be of any practical value. 21 (b) This is Wurtz reaction. In this reaction two molecules of alkyl halide react with each other to form alkane having double the number of carbon atoms. 2CH3CH2Cl + 2Na Dry ether → CH3CH2CH2CH3 + 2NaCl (X) butane ethyl chloride 22 (c) Aryl halides in presence of strong base likeNaNH2 , gives nucleophilic substitution reaction through benzyne intermediate. 23 (c) Stability of I>II hence, I is predominant. 24 (d) CCl4 is a covalent compound, therefore, it does not ionise to give Cl − ions hence, it does not give white ppt. of AgCl when treated with AgNO3 soution. There is no reaction to evolve NO2 . CCl4 will form a separate layer as it is immiscible with water. 25 (d) Nucleophilic substitution bimolecular (SN2) prefers less sterically hindered site to attack. Lesser the steric hindrance better the SN2 reaction. So, ease of reaction is 1 ° > 2 ° > 3 ° . SN2 involves inversion of configuration stereochemically (Walden inversion) 26 (a) Ethyl bromide on treating with KCN, gives ethyl cyanide, which on reduction gives propyl amine. 27 (a) CH3 − CH2 − CH2 − CH2 − Cl Alc.KOH → ∆ CH3 − CH2 − CH = CH2 + HCl 1-chlorobutane butene-1 28 (b) Alkyl halides on heating with alcoholic KOH give dehydrogenation reaction to yield alkene. If in reaction, more than one alkenes are formed, then
according to Saytzeff, the most highly substituted alkene is the major product. 29 (c) CH3Br Alc.KCN → CH3CN Na/C2H5OH → reduction CH3CH2 NH2 (X) (Y) 30 (a) Iodoform test is given by the compounds containing either | CH3CO − roup or CH3CHOH group. The structures of the given compounds are as 1. CH3CH2CH2CH2OH 2. CH3COC6H5 3. CH3CHO 4. CH3COC2H5 ∴ n butyl alcohol does not give iodoform test because it does not possess the | CH3CO − or CH3CHOH group. 31 (a) The number of monochloro derivatives of a compound depends upon the type of hydrogen present in the compound. The structure of neo- pentane is ∵It contains only one type of hydrogens. ∴ It will give only ony monochloro derivative 32 (b) Methyl alcohol (CH3OH) does not give iodoform test. 33 (a) Tertiary halide preferentially undergo SN1 substitution as they can give stable carbocation. CH3 | H3C − C − Cl Slow → −Cl (H3C)3C + +OH− → fast (H3C)3COH | carbocation t- butyl alcohol CH3 t-butyl chloride 34 (d) In [F] order of quantity of alkene 2 > 1> 3 These on addition with Br2/CCl4 to give their addition products which have C4H6 Br2 as molecular formula. (1)CH3 − CH − CH − CH3 | | Br Br Br | (2)CH3 − CH − CH − CH3 | Br (3)BrH2C − CH − CH2 − CH3 | Br (4)BrH2C − CH2 − CHBr − CH2 (5)CH2Br − CH2 − CH2 − CH2Br 35 (a) Friedel-Craft reaction : In this reaction alkyl halides react with aromatic compounds in presence of AlCl 3 or FeCl3 to form alkyl substituted aromatic compounds. 36 (a) C2H5Cl Aq.KOH → C2H5OH AgOH ← C2H5 Cl 37 (c) -OH group is converted into –Cl group by SOCl2 or anhydrous ZnCl2/conc. HCl or HCl etc. 38 (d) CHCl3 + 4NaOH → HCOONa + 3NaCl + 2H2O (aq) sodium formate 39 (a)
Among alkyl halides, iodides are least stable, hence these form Grignard reagent easily. Hence, the correct order of reactivity in formation of Grignard reagent is CH3I > CH3Br > CH3Cl 40 (a) Iodoform test is given by those compounds which have – CH3CO group or on oxidation yields this group. HCHO does not give this test. 41 (d) CH2 = CHCl + HCl → CH3 − CHCl2 ethylidene chloride sor 1, 1 dichloroethane 42 (d) Ethyl chloride can be converted into ethanol either by its alkaline hydrolysis or by its reaction with moist AgOH. C2H5Cl Aq.NaOH → ∆ C2H5OH AgOH ← C2H5Cl (A) (B) 43 (c) An organic compound forms yellow precipitate of iodoform with I2 in presence of alkali, if it has CH3CO − group directly or it has | CH3 − CHOH 44 (c) The best method for the conversion of an alcohol into an alkyl chloride is by treating the alcohol with SOCl2 in the presence of pyridine. ROH + SOCl2 → RCl + HCl + SO2 The other products being gases escape leaving behind pure alkyl halide. 45 (d) The possible isomers of hexane are Out of these structure (iii) and (i) have respectively minimum and maximum number of monochloro derivatives For structure(iii) [Only 2 monochloro derivatives (minimum) are possible] For structure (i). [5 monochloro derivatives (maximum) are possible] 47 (a) This reaction follows benzyne mechanism. 48 (a) CH3Br + OH − → CH3OH + Br − This reaction proceeds by SN2 mechanism. Rate ∝ [substrate][nucleophile] Rate ∝ [CH3Br][OH −] 49 (c) CN − (cyanide) is an ambidenate ligand, i. e., it can donate electrons to the alkyl iodide either by using carbon or by using nitrogen.