how 1-butanol may be converted to 1-pentanol-cannizzaro reaction on propanal and whyit doesnot form 1-propanol
93201
To go from 1-butanol to 1-pentanol, the easiest path to go is,\n1. Treat 1-butanol with PBr3, to yield 1-Bromobutane.\n2. Use a grignard reagent to produce CH3CH2CH2CH2MgBr.\n3. React this compound with formaldehyde, HCOH.\n4. Protonate to form 1-pentanol.\n\nCH3CH2CH2CH2-OH + PBr3 --> CH3CH2CH2CH2Br + grignard (Mg) ->\nCH3CH2CH2CH2MgBr + HCOH --> / H+ CH3CH2CH2CH2CH2-OH\n\n-----------------------------------------\nPropanal cannot undergo cannizzaro reaction because it has acidic alpha hydrogens to the carbonyl group of the aldehyde, instead it will undergo aldol condensation.\n\nhttp://en.wikipedia.org/wiki/Cannizzaro_reaction
236308
The step from pyruvate (CH3COO-) to Acetalaldehyde (CH2CHO) and CO2 is done by the enzyme pyruvate decarboxylase.  The step from Acetalaldehyde to ethanol (CH3CH2OH) is Alcohol dehydrogenase.  I do not have the chemical structures for the enzymes, because they would be way complicated, but I think what you are looking for is.\n\nC6H12O6 + 2ADP + 2 Pi = 2 CH3CH2OH + 2 CO2 + 2ATP + 2H20
688017
2-hexanol. Look at the link, I think this is what you want.
379163
(a)2-Chloro-2-methylpentane\n(b)2,3-dihydroxy-2-methylpentane\n(c)Probably 1-Ethyl-2-methylpropanesulfonic acid also can be called 2-(i-propyl)propane sulfonic acid\n(d)2,3-Epoxy-2-methylpentane (you form an epoxide of the alkene)\n(e) You'll probably get a mixture of 2,3-dibromo-2-methylpentane and 3-bromo-2-methyl-pentan-2-ol\n(f) You'll get two products.  Isopropanol and n-propanol.
959671
The rate detemining step in SN1 reactions is the first step. Step one in SN1 reaction involves carbocation formation by ionisation. Any positive feature of the alkyl halide or leaving group will increase the reaction rate:\nIn general the rate of reaction is:\nMethyl > 1° > 2° >>>3°\nSimilarly leaving group (LG) ability is a function of stability of the LG. The weaker the conjugate base, the better the LG\nFor the acids: HF, HCl, HBr, HI the acidity increases as you go down the period, so then the anion stability also increases as you go down the period. Then the general stability is:\nI- > Br- > Cl- >> F-\n\nPutting it all together you get (slowest to fastest):\nCH3Br < CH3CH2CH2I < CH3CHBrCH3 < CH3CHICH3 < (CH3)3Cl
571092
Esterification is the general name for a chemical reaction in which two chemicals (typically an alcohol and an acid) form an ester as the reaction product. These are commonly used as fragrance or flavour agents.\n\nExamples of esterification reactions include:\n\nHeating to reflux an acid (usually, but not always a carboxylic acid) and an alcohol in the presence of an catalyst forms the ester and water byproduct which can be removed to force the equibilibrium across to the desired products. This method is called Fischer esterification \nFor example, esterification of acetic acid in excess ethanol in the presence of sulfuric acid results in an ester (ethyl acetate). \nH3C-COOH + HO-CH2-CH3     →     H3C-COO-CH2-CH3 + H2O (with the presence of conc.sulfuric acid)\n\nthe reaction of an alkali carboxylate and an alcohol. For example, the reaction of sodium acetate with ethanol. \nH3C-COO- Na+ + HO-CH2-CH3 + H+ → H3C-COO-CH2-CH3 + H2O\n\nThe reaction of a silver carboxylate with a halogenoalkane (presumably driven by the insolubility of the silver halide). \nH3C-CO2-Ag+ + Cl-CH2-CH3 → H3C-COO-CH2-CH3 + AgCl\n\nThe reaction of a carboxylic acid halogenide (which is also called acyl halide) with an alcohol/phenol. \nH3C-COCl + HO-CH2-CH3 → H3C-COO-CH2-CH3 + H-Cl\n\nThe reaction of a carboxylic acid anhydride with an alcohol. (This method is used to synthesize aspirin) \nH3C-CO-O-CO-CH3 + HO-CH2-CH3 → H3C-COO-CH2-CH3 + H3C-COOH
80529
I'd expect oxidation to form the carboxylic acid. The aldehyde will form a primary carboxylic acid and CO2 gas. The ketone will form 2 secondary carboxylic acids.
1127301
I leave the H's:\nC-C-C-C-O butanol-2\nC-C-CO-C  butanol-2\nC-O-C-C-C propyloxy methane\nnot shown: isopropyloxy methane\nC-C-O-C-C ethoxy ethane
1144782
You can't have a metal alkoxide. Well, you can, but you can't. \n\nEthanol is an alcohol, and has CH3CH2OH. If a reactive metal such as cesium is mixed with it, the hydroxide ion breaks to form hydrogen gas.\n\n2CH3CH2OH (l) + Cs (s)  --> CH3CH2O- (aq) + Cs+ (aq) + H2 (g)\n\nAlkoxide is the conjugate base of the alcohol, hence CH3CH2O-
403780
GHB got a bum rap.  it's just more nonsense hysteria to justify the bloated budgets of the police state.\n\nThat aside, you don't convert beta to gamma.  suggesting an elimination failed to consider that the double bond formed will be more stable in conjugation with the carbonyl group; you'll get crotonic acid.  Subsequent hydrolysis would give you a mixture of alpha and beta.\n\nyou'd have much better luck trying to figure out a way to do it starting with GABA.\n\nOf course, if you have to ask, you lack the skilz.
813208
Isopropyl alcohol does not occur naturally.\nIt is produced from propene by indirect hydration in a\ncomplex process known as "strong-acid process", or by\ndirect catalytic reduction of acetone.
544999
==================================\nNAD+ is probably your answer.\n\nThe products of alcohol fermentation are:\nAlcohol (Ethanol) ATP, carbon dioxide \n\nSo you know for certain that your answer is either Pyruvic Acid or NAD+\n=======================\nI would argue that your teacher wants you to say NAD+ is NOT the by-product because NAD+ is converted to NADH.  Here's another way of looking at it:\n\nAlcohol fermentation is the formation of alcohol from sugar. Yeast, when under anaerobic conditions, convert glucose to pyruvic acid via the glycolysis pathways, then go one step farther, converting pyruvic acid into ethanol (alcohol).\n==============================
440879
Alcohols can be prepared by adding water to an alkene in the presence of a strong acid such as concentrated sulfuric acid. This reaction involves adding an H2O molecule across a C=C double bond. Because these reactions follow Markovnikov's rule, the product of the reaction is often a highly substituted 2 or 3 alcohol.
900980
A 1,2-diol would form.\n\nEthene would form ethan-1,2-diol, (CH2OH)2.
864248
I understand that the compound in question is soluble in both pet. ether and ethanol. The reason why pet. ether is preferred must be to do with the volatility of pet. ether. \n\nYou can easily evaporate pet. ether, which is why this is preferred over ethanol. \n\nThanks.
732185
tertiary butanol is more spherical and n-butanol is more linear. since the molecule is only polar at the hydroxyl group this is where the polar bonds are formed with water. the rest of the hydocarbon chain resists bonding and serve to decrease solubility.\n\nsince t-butanol is spherical, it has a more compact structure for its hydrocarbon portion. n-butanol on the other hand has a long, exposed alkyl chain .
425363
The initial dehydration (probably via E2 mechanism with water acting as base) forms 3-methyl-1-pentene.\n\nThat alkene is protonated by the acid present in solution (required to perform dehydration) to form a secondary carbocation.  That cation can deprotonate back to the starting alkene or to the 3-methyl-2-pentene.  The equilibrium favors the more highly substituted alkene (3-methyl-2-pentene has three alkyl groups on the double-bonded carbons).