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Begin with NCERT books, master the fundamentals. Refer to other course material and books prescribed by your teachers. Practice regularly, get difficulties solved from teachers. Spend quality time in satudy and you will succees.
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Hard work and patience will help. Have faith in your ability to crack the exam. Spend quality time in study. Practice as much as you can.Best wishes.
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Begin with NCERT books for fundamentals. After you join the coaching class refer to the books prescribed by the teachers to avoid confusion.
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The periodic table is a tabular display of the chemical elements, organized on the basis of their properties. Elements are presented in increasing atomic number. While rectangular in general outline, gaps are included in the rows or periods to keep elements with similar properties together, such as the halogens and the noble gases, in columns or groups, forming distinct rectangular areas or blocks. Because the periodic table accurately predicts the properties of various elements and the relations between properties, its use is widespread within chemistry, providing a useful framework for analysing chemical behavior, as well as in other sciences.
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Anti Markovnikov addition is a free radical reaction- proceeds through initiation, propagation and termination. Only with HBr the propagation steps are exothermic and hence peroxide effect is observed. With HCl, HF and HI one of the propagation step is endothermic, hence peroxide effect is not observed.
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Valence bond theory gives explanation for the hybridization.
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Halogens have both +R and -I effect, but the -I effect predominates over the +R effect. Therefore, all halophenols are more acidic than phenol itself. Further, since the -I effect decreases with distance, the acidic strength of halophenols decreases in the order : o-Halophenol > m-Halophenol > p-Halophenol ( where halogen = Cl, Br, I) Greater the number of electronwithdrawing groups at o- and p-positions, more stable is the phenoxide ion and hence more acidic is the phenol, theus the order is : 2,4,6-trinitro> 2,4-dinitro > 4-nitro >2-nitrophenol >phenol Formic acid > Benzoic acid > Phenol > Ethanol
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It appears that the catalytic additon of hydrogen to C=O leads to HC-OH, which then forms oxonium ion ( double boond with OH) followed by elimination of Cl and then proton establishing double bond with carbon. The other possibility is hydride ion transfer, but evidence for the same in catalytic hydrogenation is not available. The mechanism with lithium tri-tert-butoxyaluminum hydride is as follows The mechanism of the Rosenmund reduction is shown here for the reduction with lithium tri-tert-butoxyaluminum hydride. Assuming an addition-elimination mechanism, a hydride ion is transferred from lithium tri-tert-butoxyaluminum hydride to the acyl chloride in the first step. | 
| | Fig.1 | Subsequently, acid-catalyzed elimination of chloride takes place resulting in the aldehyde..
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Such formula for calculating is not known to me. For each double bond two isomers can exist- cis and trans, accordingly for each double bond you can calculate.
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The conversion can be done with Wittig Reaction. The ketone is reacted with methylenetriphenylphospharane in THF wherre C=O is replaced by C=CH2.
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Benzene on treatment with nitrating mixture of conc. HNO3 and conc. H2SO4 on heating gives nitrobenzene.
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When the two groups direct differently, i.e., one is o-p-directing and the other m-directing, then first group gets precedence. And is it reinforces each other then substitution takes place in one position only. In the given case it will be ortho wrt Cl as preferred position (i.e. position 6 on the ting) followed by position para wrt Cl (position 4 in the ring) . A group is difficult to be introduced in between the two existing groups (i.e., position 2 is difficult). Another point is both are ring deactivators so indtroducing third goup will be diffiuclt.
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Alkynes react with ammoniacal silver nitrate or cuprous nitrate to form solid product ( ppt.). Alkenes will not give this reaction.
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A solution of 100 g. (1.06 moles) of phenol in 1 l. of methanol is allowed to drop at a rate of 110 ml. per hour (Note 1) over an activated alumina catalyst (Note 2) heated to 530° (Note 3). The exit from the hot tube is attached to a receiver arranged to lead by-product gases to an efficient hood (Note 4). After addition of the methanol solution is finished, the pale yellow product is transferred to a Büchner funnel and washed with methanol. The yield of crude product melting at 135–145° is 112–115 g. (65–67%). Recrystallization from ethanol (50 g. in 650 ml.)gives 85% recovery or from benzene (50 g. in 130 ml.) gives 60% recovery of colorless hexamethylbenzene, m.p. 165–166°. 2. Notes 1. The submitters used a rate of 250 ml. per hour with a 2-in.-diameter tube, 16 in. long, packed with 300 g. of alumina, and a temperature in the catalyst bed about 370–380°. The checkers used 34 g. of alumina packed in a 7/8-in.-diameter tube 13 in. long. 2. The submitters used 4- to 8-mesh alumina from Peter Spence and Sons, Widnes, Lancashire, England. The checkers used 8- to 14-mesh Alorco H-41 obtained from the Aluminum Company of America, 1200 Alcoa Building, Pittsburgh 19, Pennsylvania. 3. Automatically controlling the outside of the catalyst tube to 370–400° gives a hot spot in the catalyst bed of 530° at the rate specified in equipment used by the checkers. |
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With available time just revise the topics atleast once and then practice solving mock tests.
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First get good medical treatment, improve your health .. With health problem perrsisting you will not be able to concentrate. Once your health improves think of exam and studies.
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Hard work and practice is necessary.
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Alkyne forms white ppt with ammoniacal silvernitrate. Alkene decolourises Br2 in CCl4 or aq. KMnO4 Alkane do not give the above reaction.
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1) Al4C3 + 12D2O ..........>3 CD4 + + 4Al(OD)3 2) Ca3N2 + H2O .........>Ca(OH)2 + NH3 3) PbS + H2O2 .......>PbSO4 + H2
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Convert acetylene to dimethyl acetylene via disodium acetylene followed by reaction with methyl bromide. Trimerization of dimethyl acetylene in presence of triethylchromium catalyst gives hexamethyl benzene. Oxidation of hexamethyl benzene with alkaline KMnO4 gives the required compound.
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