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www.chemguide.co.uk/orgmenu.html www.sparknotes.com/chemistry/
www.cem.msu.edu/virtual text Begin with NCERT books.
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Unsymmetrical breaking of a covalent bond is heterolytic fission. here the electron pair is taken away by one of the atom ( generally more electronegative) thus forms anion and the other atom loses electron hence forms cation.
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Amines can be distinguished by Hinsberg's test. In this test the amine is shaken with benzenesulphonyl chloride in presence of aq. KOH solution when i) A primary amine gives clear solution which on acidification gives insoluble N-alkylsulphonamide. ii) A secondary amine gives N,N-dialkylbenzenesulphonamide which remains unaffected on acidification. iii) A tertiary amine does not react at all, it remains insoluble in the alkaline solution.
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CIP rules are quite clear and unamibiguous. No question of any doubt in assigning priority and deciding R and S configuration. If you have any specific problem about about pseudo or real group please let me know.
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Fluorination of alkanes is too vigerous to be controlled under ordinary conditions. Fluorination brings about extensive rupture of C-C and C-H bonds leading to a mixture of products. For example, if we carry out fluorination of propane, it will form n-and isopropyl fluoride and also by braking C_C bonds will form ehtyl and methyl fluoride; also by breaking C-H bonds it will form HF.
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The o-substituents have an acid strengthening effect and this effect is unexpectedly large. Thus o-substituted benzoic acids are much stronger acids tthan their m- and p-isomers no matter whether the substituent is electron donating or electronwithdrawing. This is called as ortho effect and is combination of steric and electronic factors.
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(CH3)2CH-O-CH(CH3)2 is di-isopropyl ether
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H3PO2 is Reducing agent, as it will itself undergo oxidation and reduce the substitute compound/ element
Now in H3PO2.......all the hydrogen atoms are not directly connected with phosphorus atom...thus when energy provide, resonance takes place and 1 or 2 hydrogen atom will be donated by it.........thus it is a reducing agent
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The product formed would be (CH3)2C=CH-CH3. The reaction cannot proceed by E2 mechanism as the beta carbon is hindered and has no hydrogen. Thus it will follow E1 mechanism proceeding though carbo cation. The initially formed neopentyl carbocation [CH3-C(CH3)2CH2+] being primary is less stable it will rearrange to more stable carbocation by migration of methyl group leading to tertiary carbocation which is more stable (CH3)2C+-CH2CH3. this will then lose a proton to give the final product.
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Fehling's solution test is given by aldehydes only.
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Benzene is rich in electrons hence gives electrophilic substitution reactions. An electrophile attacks benzene ring to form resonance stabilized carbocation which eliminates a proton / loses a proton to base formed in the first step while generating electrophile, to give the substitution product. Please refer to following for details. www.ch.ic.ac.uk/local/organic/tutorial/EHS_1.pdf
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I have not noticed any differrence. Its a good reference book. Use Finar for reactions.
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Definition: A value used to indicate the resistance of a motor fuel to knock. Octane numbers are based on a scale on which isooctane is 100 (minimal knock) and heptane is 0 (bad knock). Examples: A gasoline with an octane number of 92 has the same knock as a mixture of 92% isooctaneand 8% heptane.
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Phenol on losing a proton forms phenoxide ion which is more stable than phenol itself due to resonance, thus phenol easily loses a proton and is acidic in nature. Alkoxide ion is not stable than alcohol hence alcohol does not lose proton easily, and is not acidic.
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Bromoethane (C2H5Br) can be prepared by reacting ethane with bromine at 520K. If you need bromoethene (BrCH=CH2), then the above bromoethane is converted to ethene (CH2=CH2) by the action of alcoholic KOH, which is then converted to dibromoethane (BrCH2-CH2Br), this on reaction with alc.KOH first followed by NaNH2 in liq. NH3 gives acetylene, further addition of HBr to acetylene gives bromoethene.
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SN1 reaction proceeds through formation of carbocation. The C6H5 group can stabilize the carbocation by resoance, thus more the number of C6H5 groups more stable will be the carbocation.
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CCl3 would act like a ring deactivator and meta director group. The substitution in the ring will take place at m-position wrt to CCl3. The same electrophilic substitution mechanism.
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During reductive ozonolysis the C=C breaks and is converted to CHO. Thus in cyclopentene also the double bond will break and the carbons forming the double bond get converted to CHO, thus the product should be CHO-CH2-CH2-CH2-CHO. The question of ring being again formed should not arise. When the above product is treated wtih dil NaOH , it can undergo internal aldol condensation and form the four membered ring , 2-hydroxycyclobutane-1-carbaldehyde.
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I am pasting some examples which should solve your difficulty. | | | Examples of Steric Inhibition of Resonance 
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Which one will be more acidic ? Because of steric inhibition of resonance conjugate base of I will not be stabilised by resonance. But for II there is no such steric inhibition of resonance.
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Its good to start early to win a race.. Begin with NCERT books, if fundamentals are clear then it becomes easier to understand the subjects.
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