1LOLA
1. Why are alkynes less reactive towards electrophilic reagents like H+ than alkenes???
2. How to prepare benzene from lime?
3. What is chloral?
4.When gas A is passed through dry KOH at low temperature, a deep red coloured compound B and gas C are obtained. The gas on treatment with but-2-ene followed by treatment with Zn/H2 yeilds acetaldehyde. Identify A, B and C.
5. What is lucas reagent?
6. On monochlorination of 2-methylbutane, how many enantiomeric pairs are formed?
7. arrange according to rate of hydrolysis...
CH3COOC2H5 C2H5COCl (CH3CO)2O CH3CONH2
8. Show all steps of reaction when we add AgNO3/NH4OH and P2O5 such that final product is propionic anhydride.
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27 Answers
39Bond dissociation energy represents homolytic fission in any case. It does not depict the energy of a heterolytic bond fission, and hence cannot say anything about acidity or basicity. Thus BDE reflects stability of the radical so formed after homolytic cleavage.
We know iodine radicals are unstable(like I just said in my previous post), thus if they have similar BDEs, fluorine radicals must be unstable too (and not without reason...we all know how explosive/uncontrolled/extremely reactive/blah blah fluorine is).
About BDEs, read this : http://masterorganicchemistry.com/2010/06/25/bond-dissociation-energies-homolytic-cleavage/
71One Question from my side:
F2 and I2 have nearly same standard Homolytic Bond dissociation energy. Please Explain why?
H° (F2) = 159 kJ/mol
H° (I2) = 151 kJ/mol
39Because the bond enthalpy of the H-Cl bond is too high to break under Kharash effect conditions (the bond is strongly ionic, homolytic fission is not possible) and for HI, iodine can form radicals but as per Reactivity-Selectivity principle, iodine radicals are unstable and combine to form iodine molecules instead.
1why does peroxide effect acts only on HBr but not on HCl or HI?
39Most of them have been answered....
8) We react propanal with Tollen's reagent, which oxidizes the aldehyde to propanoic acid. Then with 2 mols of acid, we use P2O5 as a dehydrating agent and our product is propionic anhydride.
49actually i was getting confused over some basic things so wanted to clear them off ! :P
any ways all these are extracts from IITJEE papers! :)
61) consult NCERT.............coz they ahv a cylindrical field of delocalized pi electrons.....this is only a hint:P)
2) i too think it shud be soda lime(coz its easier in that case:P)
3) CCl3 CHO
5) Lucas' reagent is a solution of zinc chloride in concentrated hydrochloric acid.
1 more thing , how much portion of chem hav u completed subho??
these r pretty easy ones !!:D
110 i m not sure
CH3CH2OH in PCC -> CH3CHO
CH3CH2OH with K2Cr2O7 -> CH3COOH
CH3COOH + CH3CHO -> CH3COOCH(OH)CH3
CH3COOCH(OH)CH3 -> in ThO2 -> CH3COOCH=CH2
i m not sure abt whether ThO2 will work as reqd in last step may be u can figure out how to dehydrate in the last step cheers
8 given reagents are added to wat ?
17th is it b>c>a>d
2nd CaO +coke gives CaC2 (calcium carbide )
CaC2 on hydrolysis will give acetylene, pass it through red hot Fe tube u will get benzene
49add more..
9. Iodoform is obtained from reaction of acetone with hypoiodite but not with iodide. Why??
10. Convert ethyl alcohol to vinyl acetate in not more than 6 steps.
496) we are generally getting 1 enantiomeric pair but the answer given is 2 enantiomeric pairs.
399)
To prepare idoform m, I+ion is required which is supplied by IO- which iodine cant help in
394)
http://books.google.com/books?id=93IGOHrhpZUC&pg=SA3-PA97&lpg=SA3-PA97&dq=When+gas+A+is+passed+through+dry+KOH+at+low+temperature,+a+deep+red+coloured+compound+B+and++gas+C+are+obtained.+The+gas+on+treatment+with+but-2-ene+followed+by+treatment+with+Zn/H2+yeilds+acetaldehyde.+Identify+A,+B+and+C.&source=bl&ots=AfOhq4Z5xX&sig=htiSBFLdvKebW5hmkIHXEBul8Yk&hl=en&ei=_N46Td-qMo3krAfiiIm7CA&sa=X&oi=book_result&ct=result&resnum=6&ved=0CCMQ6AEwBQ#v=onepage&q&f=false
396) u decide form this:
the products are:
1)1-chloro-2-methylbutane
2)2-chloro-2-methylbutane
3)3-chloro-2-methylbutane
4)4-chloro-2-methylbutane
391)
not exactly sure if this contains the answer
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/addyne1.htm
39u cant use lucas test to differentiate between tertiary and allylic, benzylic alcohols
395)
http://www.goiit.com/posts/list/organic-chemistry-what-is-the-relationship-between-lucas-test-1015678.htm;jsessionid=0E264DBB2B73AC3C222E9B71EEA0849A#1262326
395)
Lucas test in alcohols is a test to differentiate between primary,secondary and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides.File:Lucas test.png
When Lucas' reagent (ZnCl2 in concentrated HCl solution) is added to the alcohol, H+ from HCl will protonate the -OH group of alcohol, so that the leaving group H2O, being a much weaker nucleophile than OH-, can be substituted by nucleophile Cl-. Lucas' reagent offers a polar medium in which SN1 mechanism is favored. In unimolecular nucleophilic substitution, the reaction rate is faster when the carbocation intermediate is more stabilized by greater number of electron donating alkys group (R-) bonded to the positively charged carbon atom. Tertiary alcohols react immediately with Lucas reagent to produce turbidity while secondary alcohols do so in five minutes. Primary alcohols do not react appreciably with Lucas reagent at room temperature.
The reagent dissolves the alcohol, removing the OH group, forming a carbocation. The speed of this reaction is proportional to the energy required to form the carbocation, so tertiary, benzylic, and allylic carbocations react quickly, while smaller, less substituted, alcohols react more slowly. The cloudiness observed is caused by the carbocation immediately reacting with the chloride ion creating an insoluble chloroalkane.