15. Halogen compounds
A section of Chemistry, 9701
Listing 10 of 284 questions
The hydrocarbons A, C4H10, and B, C4H8, are both unbranched. A does not decolourise bromine. B decolourises bromine and shows geometrical isomerism. Draw the skeletal formula of A. A The hydrocarbon A, C4H10, has a branched isomer. Suggest why unbranched A has a higher boiling point than its branched isomer. Give the structural formula of B. Explain why B shows geometrical isomerism. Draw the mechanism of the reaction of B with bromine, Br2. Include all necessary charges, dipoles, lone pairs and curly arrows. Explain the origin of the dipole on Br2 in this mechanism. The alcohols C and D are isomers of each other with molecular formula C4H10O. Both isomers are branched. When C is heated under reflux with acidified potassium dichromate(no colour change is observed. When D is heated under reflux with acidified potassium dichromate(the colour of the mixture changes from orange to green and E, C4H8O2, is produced. E reacts with aqueous sodium carbonate to form carbon dioxide gas. Identify C, D and E. C D E Write the equation for the reaction between E and aqueous sodium carbonate. The isomers F and G, C5H10O, both form an orange precipitate when reacted with 2,4-DNPH. F is unbranched and reacts with alkaline aqueous iodine to produce a yellow precipitate. G does not react with alkaline aqueous iodine. It contains a chiral centre and produces a silver mirror when warmed with Tollens’ reagent. Name the yellow precipitate produced by the reaction between F and alkaline aqueous iodine. Give the structural formula of F and of G. F G Explain the meaning of the term chiral centre. H and I are isomers with molecular formula C2H4O2. The infra-red spectra of isomers H and I are shown. percentage transmittance percentage transmittance wavenumber / cm–1 H wavenumber / cm–1 I Identify the bonds responsible for the principal peaks above 1500 cm–1 in each spectrum. spectrum of H spectrum of I Name H and I. H I
9701_s17_qp_22
THEORY
2017
Paper 2, Variant 2
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Compounds J, K, L and M are isomers of each other with the molecular formula C9H11NO. All four isomers contain a benzene ring. Two of the isomers contain a chiral centre. The results of six tests carried out on J, K, L and M are shown in the table. test observations with each isomer J K L M add cold HCl soluble soluble soluble insoluble add 2,4-DNPH reagent orange ppt. orange ppt. orange ppt. no reaction add NaOH+ I2pale yellow ppt. no reaction pale yellow ppt. no reaction warm with Fehling’s solution no reaction red ppt. no reaction no reaction heat with NaOHno reaction no reaction no reaction P(C6H7N) and Q(C3H5O2Na) produced diazotization and addition of alkaline phenol no dye produced orange dye produced no dye produced no dye produced Use the experimental results in the table above to determine the group, in addition to the benzene ring, present in each of the four isomers J, K, L and M. Complete the table below, identifying the grouppresent in each isomer. groupin compound J K L M Name the type of reaction occurring in test 5 that converts M into P + Q. Suggest structures for compounds P and Q. P (C6H7N) Q (C3H5O2Na) Isomers J, K, L and M all have the molecular formula C9H11NO. Use the information in to suggest a structure for each of these isomers and draw these in the boxes. Draw circles around all chiral centres in K and L. J K L M Compound N is another isomer which has the same molecular formula C9H11NO and also contains a benzene ring. N contains the same functional group as M. When heated with NaOH, N produces ethylamine and a sodium salt W. Suggest the structure of W. W
9701_s17_qp_41
THEORY
2017
Paper 4, Variant 1
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Compounds J, K, L and M are isomers of each other with the molecular formula C9H11NO. All four isomers contain a benzene ring. Two of the isomers contain a chiral centre. The results of six tests carried out on J, K, L and M are shown in the table. test observations with each isomer J K L M add cold HCl soluble soluble soluble insoluble add 2,4-DNPH reagent orange ppt. orange ppt. orange ppt. no reaction add NaOH+ I2pale yellow ppt. no reaction pale yellow ppt. no reaction warm with Fehling’s solution no reaction red ppt. no reaction no reaction heat with NaOHno reaction no reaction no reaction P(C6H7N) and Q(C3H5O2Na) produced diazotization and addition of alkaline phenol no dye produced orange dye produced no dye produced no dye produced Use the experimental results in the table above to determine the group, in addition to the benzene ring, present in each of the four isomers J, K, L and M. Complete the table below, identifying the grouppresent in each isomer. groupin compound J K L M Name the type of reaction occurring in test 5 that converts M into P + Q. Suggest structures for compounds P and Q. P (C6H7N) Q (C3H5O2Na) Isomers J, K, L and M all have the molecular formula C9H11NO. Use the information in to suggest a structure for each of these isomers and draw these in the boxes. Draw circles around all chiral centres in K and L. J K L M Compound N is another isomer which has the same molecular formula C9H11NO and also contains a benzene ring. N contains the same functional group as M. When heated with NaOH, N produces ethylamine and a sodium salt W. Suggest the structure of W. W
9701_s17_qp_43
THEORY
2017
Paper 4, Variant 3
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X is CH3CH(OH)CH2CH3. The reaction between X and alkaline aqueous iodine produces a yellow precipitate. Give the name of the compound formed as a yellow precipitate in this reaction. Give the name of X. There are three structural isomers of X that are alcohols. Draw the structures of these three isomers of X.  Two reactions of X are shown. CH3CH(OH)CH2CH3 CH3COCH2CH3 C4H8 reaction 2 reaction 1 X Identify the type of reaction involved in reaction1. Identify the reagents for reaction1. Reaction2 can be carried out by passing the vapour of X over hot aluminiumoxide. The product of reaction2, C4H8, is actually a mixture of three isomers. Give the full names of the three isomers formed by reaction2.  The reaction of methylpropene, (CH3)2CCH2, with hydrogenbromide, HBr, produces a mixture of two halogenoalkanes. One of the halogenoalkanes, 2‑bromo‑2‑methylpropane, is formed as the major product while 1‑bromo‑2‑methylpropane is formed in small quantities. Complete the mechanism to show the reaction of methylpropene with HBr to form the major product. Include the structure of the intermediate and all necessary charges, dipoles, lone pairs and curly arrows. The structure of 2‑bromo‑2‑methylpropane is not required. CH3 H3C C H H C H Br 2-bromo-2-methylpropane  Explain why 2-bromo-2-methylpropane is the major product of this reaction. 
9701_s18_qp_21
THEORY
2018
Paper 2, Variant 1
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A is CH3CHBrCH2CH3. Some reactions of A are shown. CH3CHBrCH2CH3 H+ / Cr2O7 2– CH3CH(OH)CH2CH3 C4H8 A C4H8O B reaction 2 NaOH reaction 1 NaOH Name A. Name the class of compound to which B belongs. There are three structural isomers of A. Draw the structures of these three isomers of A.  Reaction1 occurs by two different mechanisms at the same time. These mechanisms are referred to as SN1 and SN2. State what the letters ‘S’ and ‘N’ represent in the abbreviation SN1. S N  Complete the SN1 mechanism for reaction1. Include the structure of the intermediate and all necessary charges, dipoles, lone pairs and curly arrows. CH3 H3C CH2 C Br H CH3 H3C CH2 C OH H  The SN1 mechanism for reaction1 is repeated using CH3CHCl CH2CH3 or CH3CHICH2CH3 in place of the CH3CHBrCH2CH3. State and explain how the rates of these two reactions will compare with the rate of the original reaction using CH3CHBrCH2CH3. Reaction2 uses the same reagent as reaction1, but under different conditions. State two differences in the conditions needed to ensure that reaction2 is more likely to take place than reaction1 when this reagent is added. 
9701_s18_qp_23
THEORY
2018
Paper 2, Variant 3
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